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A 2.1 -g Ping-Pong ball is rubbed with a piece of wool, resulting in a net positive charge of \(7.4 \mu \mathrm{C}\) on the ball. How many electrons were transferred from the ball to the wool in this process?

Short Answer

Expert verified
Approximately \(4.6 \times 10^{13}\) electrons were transferred from the ball to the wool.

Step by step solution

01

Understand Charge Units

The problem states that the ball has a net positive charge of \(7.4 \mu \mathrm{C}\), which is 7.4 microcoulombs. A microcoulomb is \(1 \times 10^{-6}\) coulombs, so \(7.4 \mu \mathrm{C} = 7.4 \times 10^{-6}\, C\).
02

Know Elementary Charge

The elementary charge (charge of one electron) is \(-1.6 \times 10^{-19}\, C\). Although electrons have a negative charge, we are looking for the number moved, so we'll use the absolute value.
03

Calculate Number of Electrons Transferred

To find the number of electrons transferred, we divide the total charge in coulombs by the magnitude of the charge on a single electron: \[\text{Number of electrons} = \frac{7.4 \times 10^{-6}\, C}{1.6 \times 10^{-19}\, C/e^-}\] Performing the calculation gives us \(4.625 \times 10^{13}\, \text{electrons}\).
04

Convert and Round the Answer

The calculation results in \(4.625 \times 10^{13}\) electrons. Typically, we round this to an appropriate significant figure based on the precision of the given charge value. Here, we will round to \(4.6 \times 10^{13}\) electrons.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Charge Transfer
When we talk about charge transfer in electrostatics, we are referring to the movement of electrical charge between two objects. In the case of the Ping-Pong ball and the wool, rubbing them together causes electrons to move from one to the other. This transfer is what results in the net charge we observe.
The charge transfer occurs because the electrons are physically moved due to the force of friction when the surfaces are in contact. The Ping-Pong ball gains or loses electrons depending on which object's material has a greater affinity for electrons.
The key takeaway is that charge transfer does not involve a change in the total number of charges but a redistribution. One object loses electrons while the other gains them. The total electric charge is conserved in an isolated system during the transfer.
Elementary Charge
The elementary charge is the charge of a single electron or proton, and it is fundamental in electrostatics. This is the smallest unit of charge that is considered indivisible in classical physics. The value of the elementary charge is approximately \\(-1.6 \times 10^{-19} C\) for an electron and \\(+1.6 \times 10^{-19} C\) for a proton.
In charge transfer calculations, we often use the magnitude of the elementary charge to determine the number of electrons transferred. Even though electrons themselves have a negative charge, the focus in calculations is on the quantity transferred, for which we take the absolute value.
This standard measurement is crucial for any calculation involving electric charge, whether it's in a classroom exercise or real-world applications, ensuring consistency and accuracy in understanding electrical phenomena.
Electron Count
Counting electrons in a process that involves charge transfer is key to understanding the extent of that transfer. By calculating how many electrons move from one object to another, we get a clearer picture of how positive or negative a net charge becomes.
In our example, to find the electron count, we need to divide the total charge transferred (in coulombs) by the elementary charge. This formula: \\[ \text{Number of electrons} = \frac{\text{Total Charge}}{\text{Elementary Charge}} \]
allows us to understand how even tiny charges can involve large numbers of electrons due to their incredibly small individual charge. With practice, this calculation becomes intuitive, illustrating why small forces generate noticeable effects at a macroscopic level.
Unit Conversion
Unit conversion is a critical skill in physics, ensuring all measurements used in calculations are in compatible units. For charge transfer problems, expressed values must be converted into the same units to perform calculations correctly.
In the example problem, a net charge was initially given in microcoulombs. Since most charge calculations require units in coulombs, converting microcoulombs to coulombs by multiplying by \\(1 \times 10^{-6}\) ensures accuracy. \So, \\(7.4 \mu \text{C} = 7.4 \times 10^{-6} \text{C}\).
Applying this conversion technique allows for seamless calculations and deeper insights into electrical interactions. Regardless of the discipline, mastering unit conversions is foundational for solving complex scientific and engineering problems, allowing for clarity and precision in every step.

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Most popular questions from this chapter

A point charge \(+6.5 \mu C\) is placed at the origin, and a second charge \(-4.2 \mu \mathrm{C}\) is placed in the \(x-y\) plane at \((0.35 \mathrm{~m}, 0.45 \mathrm{~m})\). Where should a third charge be placed in the plane so that the net force acting on it is zero?

The net charge of one mole of protons is (a) \(1.60 \times 10^{-19} \mathrm{C}\); (b) \(1.00 \mathrm{C} ;\) (c) \(1.60 \times 10^{4} \mathrm{C} ;\) (d) \(9.65 \times 10^{4} \mathrm{C}\)

A proton moving at \(3.8 \times 10^{5} \mathrm{~m} / \mathrm{s}\) to the right enters a region where a uniform electric field of magnitude \(56 \mathrm{kN} / \mathrm{C}\) points to the left. (a) How far will the proton travel before it stops moving? (b) What will happen to the proton after it stops moving? (c) Graph the proton's position versus time, velocity versus time, and acceleration versus time after it enters the field.

Two parallel plates carry uniform charge densities \(-0.50 \mathrm{nC} / \mathrm{m}^{2}\) and \(+0.50 \mathrm{nC} / \mathrm{m}^{2}\). (a) Find the electric field between the plates. (b) Find the acceleration of an electron between these plates.

Electric bees. As they fly through the air, honeybees can pick up a small amount of electric charge. Suppose a \(0.12-\mathrm{g}\) bee acquires a charge of \(-55 \mathrm{pC}\). (a) How many excess electrons has the bee acquired? (b) Find the electric force on the bee in the electric field just above Earth's surface, \(120 \mathrm{~N} / \mathrm{C}\) directed vertically downward. (c) Compare your answer in part (b) to the bee's weight.

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